Question

Answer 1

$$\begin{gathered} \int \left(\frac{1+\cos 4x}{\cot x-\tan x}\right)dx \\ =\int \frac{\left(1+\cos 4x\right)}{\left({\displaystyle \frac{\cos x}{\sin x}}-{\displaystyle \frac{\sin x}{\cos x}}\right)}dx \\ =\int \frac{2{\cos }^{2}2x\times \sin x\cos x}{\left({\cos }^{2}x-{\sin }^{2}x\right)}dx \\ =\int \frac{{\cos }^{2}2x\times 2\sin x\cos x}{\cos 2x}dx \\ =\int \cos 2x\sin 2xdx \\ =\frac{1}{2}\int 2\sin 2x\cos 2xdx \\ =\frac{1}{2}\int \sin 4xdx \\ =\frac{1}{2}\left[-\frac{\cos 4x}{4}\right]+\mathrm{C} \\ =-\frac{1}{8}\cos 4x+\mathrm{C} \end{gathered}$$